Physics Olympiad 1999 Q3

noospace
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Physics Olympiad 1999 Q3

Postby noospace » Mon Sep 28, 2009 8:31 pm

http://www.princeton.edu/~jdpeters/docs ... piad99.pdf

3. A uniform 2 kg cylinder rests on a lab cart as shown. The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cm in diameter and 10 cm high, which of the following is closest to the minimum acceleration required to cause the cylinder to tip over?

This question had me a bit confused for a while. I found that I could get the right answer by assuming the static friction is essentially infinite and then transforming into the cart frame where the cylinder experiences an inertial force of m*a in addition to gravity m*g acting down. Then applying torque balance about the lower corner of the cylinder gives me

5 cm * m *a = 2 cm * m * g

and thus a = 0.4* g = 4 m/s/s

Does anyone know of a more elegant approach to this question?

Thanks

blackcat007
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Re: Physics Olympiad 1999 Q3

Postby blackcat007 » Mon Sep 28, 2009 10:12 pm

noospace wrote:http://www.princeton.edu/~jdpeters/docs/physics_olympiad99.pdf

3. A uniform 2 kg cylinder rests on a lab cart as shown. The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cm in diameter and 10 cm high, which of the following is closest to the minimum acceleration required to cause the cylinder to tip over?

This question had me a bit confused for a while. I found that I could get the right answer by assuming the static friction is essentially infinite and then transforming into the cart frame where the cylinder experiences an inertial force of m*a in addition to gravity m*g acting down. Then applying torque balance about the lower corner of the cylinder gives me

5 cm * m *a = 2 cm * m * g

and thus a = 0.4* g = 4 m/s/s

Does anyone know of a more elegant approach to this question?

Thanks

moment balancing is the only condition for an eqm. thus i think this is the only pith of the problem..

noospace
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Re: Physics Olympiad 1999 Q3

Postby noospace » Mon Sep 28, 2009 11:56 pm

Yeah, but one would think that there's a better way than considering the moments of fictitious forces.

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WhoaNonstop
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Re: Physics Olympiad 1999 Q3

Postby WhoaNonstop » Tue Sep 29, 2009 12:03 am

Well, you can't really "exactly" solve the problem, and by viewing the other problems on the test, it appears that there isn't can't be much to it.

How I would go about "reasoning" the answer would be through F = ma.

Where, the force on the cylinder, the friction is uFn = ma.
Also Fn - mg = 0 (Fn = mg) and therefore ug = a.

Solving of course, a = 4.9 m/s^2.

However, since a force is exerted on the top end of the cylinder, the drag force, you could safely assume that it would fall over at less than 4.9 m/s^2. So 4.0 m/s^2 becomes the most reasonable choice.

I am not sure of an exact way to do it myself. It seems like there would have to be more info. =)

-Riley

noospace
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Re: Physics Olympiad 1999 Q3

Postby noospace » Tue Sep 29, 2009 7:12 pm

Hi WhoaNonstop,

No. I think this is a torque balancing problem as stated earlier.

michael
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Re: Physics Olympiad 1999 Q3

Postby michael » Thu Oct 22, 2009 7:20 am

Hey,
I agree that the solution is not obvious to this problem. But I think the following method is fairly rigorous:

Consider moments about the COM of the cylinder - (correct me if I'm wrong, but we can always choose any point we like to take moments about -right?)

There are 3 forces on the block:
1) mg going down through the COM
2) a reaction force R, which when the cylinder is just about to tip, is upward, and equal to mg THROUGH THE BACK CORNER OF THE BLOCK, NOT THROUGH THE COM
3) a frictional force (F) in the direction of motion of the cart on the base of the cylinder

The maximum friction F_max that can be supplied to the cylinder is mu R = mu mg. For accelerations a, the force F will be:
m a for m a < F_max
F_max for m a > F_max

The first thing to check is that the max force is large enough to tip the cylinder (height h and diameter d):
m mu g (h/2) > mg (d/2) is equivalent to mu h > d which is true since 0.5* 10 = 5 > 4 so the block can tip and (E) is ruled out

Now check what min force causes tipping. When just about to tip, the condition of moment balance occurs:
m a_tip (h/2) = m g (d/2) thus a_tip = g(d/h) = 4m/s^2

I think the most dodgy bit of this argument is that mu would be different when all the force was acting through a point (i.e. just before tipping) than when spread over the surface. But I doubt they would consider those sorts of effects in this kind of a multiple choice test!




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